$f(n) = 2n$ $h(x) = -6x^{2}+2x-f(x)$ $g(x) = 5x+2-3(f(x))$ $ h(g(7)) = {?} $
Answer: First, let's solve for the value of the inner function, $g(7)$ . Then we'll know what to plug into the outer function. $g(7) = (5)(7)+2-3(f(7))$ To solve for the value of $g$ , we need to solve for the value of $f(7)$ $f(7) = (2)(7)$ $f(7) = 14$ That means $g(7) = (5)(7)+2+(-3)(14)$ $g(7) = -5$ Now we know that $g(7) = -5$ . Let's solve for $h(g(7))$ , which is $h(-5)$ $h(-5) = -6(-5)^{2}+(2)(-5)-f(-5)$ To solve for the value of $h$ , we need to solve for the value of $f(-5)$ $f(-5) = (2)(-5)$ $f(-5) = -10$ That means $h(-5) = -6(-5)^{2}+(2)(-5)-(-10)$ $h(-5) = -150$